gpt pro 5.2 attempt on Math-5608

Let $\varepsilon:=a_1+b_1\sqrt d>1$. Then $\varepsilon^n$ has norm $1$ and equals $a_n+b_n\sqrt d$ with integers $a_n,b_n$ (by repeated multiplication), so each $n$ yields a solution.

Conversely, let $u=x+y\sqrt d>1$ be any solution. Choose $n\in\mathbb{Z}$ such that $\varepsilon^n\le u<\varepsilon^{n+1}$. Then $v:=u/\varepsilon^n$ is again a unit of norm $1$, and satisfies $1\le v<\varepsilon$. Writing $v=r+s\sqrt d$ with $r,s\in\mathbb{Z}$ (closure under multiplication and inversion in this unit group), the inequalities $1\le v<\varepsilon$ force $s=0$ and $r=1$ (since $|r-s\sqrt d|=|\overline v|=1/v\le 1$ and $r,s$ are integers). Hence $v=1$ and $u=\varepsilon^n$. Therefore every solution is a power of the fundamental unit. ∎

Step 1: basic coprimality. From $a_1^2-db_1^2=1$, any common divisor of $a_1$ and $d$ must divide $1$, so $\gcd(a_1,d)=1$ and in particular $a_1>1$.

Step 2: divisibility in the $b_n$ sequence. For $n=kl$ we have \[ a_{kl}+b_{kl}\sqrt d = (a_k+b_k\sqrt d)^l. \] Expanding the right-hand side by the binomial theorem, every term contributing to the $\sqrt d$-coefficient contains a factor $b_k$, so $b_k\mid b_{kl}$. In particular, taking $k=1$ shows $b_1\mid b_n$ for all $n$, so we may write \[ b_n=b_1c_n\qquad (c_n\in\mathbb{N}). \] Moreover, for any $l\ge 1$, dividing the binomial expansion for $(a_1+b_1\sqrt d)^l$ by $b_1$ yields the explicit identity \[ c_l = l a_1^{l-1} + \binom{l}{3} a_1^{l-3} b_1^2 d + \binom{l}{5} a_1^{l-5} b_1^4 d^2 + \cdots, \tag{1} \] a sum over odd indices.

Step 3: prime divisors of $m$ lead to contradiction. Assume for contradiction that $m>1$. Let $p$ be any prime divisor of $m$, and write $m=pr$. Then $b_p\mid b_m$, hence $c_p=b_p/b_1\mid b_m$. Since $b_m$ is a $d$-number, every prime factor of $c_p$ divides $d$, i.e. $c_p$ is a $d$-number. We now show this is impossible for any prime $p$.

Case $p=2$. From $1$ with $l=2$ we have $c_2=2a_1$. If $c_2$ is a $d$-number, every prime factor of $a_1$ divides $d$, contradicting $\gcd(a_1,d)=1$ unless $a_1=1$. But $a_1>1$, contradiction.

Case $p=3$. From $1$ with $l=3$ we get \[ c_3 = 3a_1^2 + b_1^2 d. \] Using $a_1^2-db_1^2=1$ we rewrite $b_1^2 d=a_1^2-1$, hence \[ c_3 = 4a_1^2-1=(2a_1-1)(2a_1+1). \] Let $q$ be any prime divisor of $c_3$. Since $c_3$ is a $d$-number, $q\mid d$. Reducing $c_3=3a_1^2+b_1^2d$ modulo $q$ gives $0\equiv 3a_1^2\pmod q$. As $\gcd(a_1,q)=1$ (because $q\mid d$ and $\gcd(a_1,d)=1$), we conclude $q=3$. So $c_3$ is a power of $3$, meaning both $2a_1-1$ and $2a_1+1$ are powers of $3$ differing by $2$. The only positive powers of $3$ differing by $2$ are $1$ and $3$, which forces $a_1=1$, contradiction.

Case $p\ge 5$. Let $q$ be any prime divisor of $c_p$. Again $c_p$ is a $d$-number, so $q\mid d$. Reduce identity $1$ modulo $q$ with $l=p$: all terms after the first contain a factor $d$, hence vanish modulo $q$, so \[ 0\equiv c_p \equiv p a_1^{p-1}\pmod q. \] Since $\gcd(a_1,q)=1$, this implies $q\mid p$, so $q=p$. Thus every prime factor of $c_p$ is $p$, so $c_p=p^r$ for some $r\ge 1$. In particular $p\mid c_p$, hence $p\mid d$. Now in $1$ with $l=p$, every term after the first is divisible by $p^2$ because it has a factor $d$ (hence $p$) and a binomial coefficient $\binom{p}{j}$ (hence $p$). Therefore $c_p > p a_1^{p-1}$ (since all summands are positive), and because $a_1\ge 2$ we have $p a_1^{p-1}\ge p\cdot 2^{p-1} > p$. So $c_p>p$, contradicting $c_p=p^r$ with $r=1$, and also contradicting $r\ge 2$ because then $c_p\ge p^2$ would force $p^2\mid p a_1^{p-1}$ and hence $p\mid a_1$, again contradicting $\gcd(a_1,d)=1$.

In all cases we obtain a contradiction, so $m$ has no prime divisors, hence $m=1$. Thus the only solution with $y$ a $d$-number is the fundamental one, proving uniqueness. ∎

Let $y$ be $S$-smooth and set $a:=2y+1$. If also $y+1$ is $S$-smooth then $4y(y+1)$ is $S$-smooth, and \[ 4y(y+1)=(2y+1)^2-1=a^2-1. \] So it suffices to bound the number of $S$-smooth integers of the form $a^2-1$.

Write $a^2-1=\prod_{i=1}^r p_i^{e_i}$. Define integers $b_i\ge 0$ by:

• if $e_i$ is odd, set $b_i:=e_i-1$ (so $e_i-b_i=1$);
• if $e_i\in\{0,2\}$, set $b_i:=0$ (so $e_i-b_i\in\{0,2\}$);
• if $e_i\ge 4$ is even, set $b_i:=e_i-2$ (so $e_i-b_i=2$).

For each fixed $d$, Lemma 2.2 shows there is at most one positive solution $(a,b)$ to $a^2-d b^2=1$ with $b$ a $d$-number. Hence there are at most $3^r$ possible values of $a$, and therefore at most $3^r$ possible values of $y=(a-1)/2$. This proves finiteness. ∎